60.0k views
5 votes
A 4-A current is maintained in a simple circuit with a total resistance of 2 Ω. How much energy is dissipated in 3 seconds?

A) 3 J
B) 6 J
C) 12 J
D) 24 J
E) 96 J

User Mkko
by
4.8k points

1 Answer

1 vote

Answer: Energy dissipated E = 96J

Step-by-step explanation:

Given:

Current I = 4A

Resistance R = 2 Ohms

Time t = 3 seconds

The energy dissipation in an electric circuit can be derived from the equation below:

E = IVt ....1

Where;

I = current, V = Voltage (potential difference), t= time and E = energy dissipated

But we know that;

V = I×R .....2

Substituting equation 2 to 1, we have

E = IVt = I(I×R)t = I^2(Rt)

Substituting the values of I,R and t

E = 4^2 × 2 ×3 = 96J

Energy dissipated E = 96J

User Jeremy Goodell
by
5.5k points